A374384 a(n) = floor(Sum_{k=n^3..(n+1)^3} k^(1/3)).
1, 12, 51, 134, 281, 508, 835, 1278, 1857, 2588, 3491, 4582, 5881, 7404, 9171, 11198, 13505, 16108, 19027, 22278, 25881, 29852, 34211, 38974, 44161, 49788, 55875, 62438, 69497, 77068, 85171, 93822, 103041, 112844, 123251, 134278, 145945, 158268, 171267, 184958, 199361
Offset: 0
Links
- Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).
Programs
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Mathematica
Table[Floor[Sum[(n^3+k)^(1/3),{k,0,3n^2+3n+1}]],{n,0,40}] (* Stefano Spezia, Jul 07 2024 *) -
PARI
a(n) = 3*n^3+9*n^2\2+4*n+1; \\ Michel Marcus, Jul 09 2024
Formula
a(n) = floor(3*n^3+9*n^2/2+4*n+1).
a(2*n) = 24*n^3 + 18*n^2 + 8*n + 1.
a(2*n-1) = 24*n^3-18*n^2+8*n-2 for n > 0.
a(2*n) = A248575(2*n) + 4*n + 1.
a(2*n-1) = A248575(2*n-1) + 4*n - 2.
From Stefano Spezia, Jul 09 2024: (Start)
G.f.: (1 + 9*x + 17*x^2 + 7*x^3 + 2*x^3)/((1 - x)^4*(1 + x)).
E.g.f.: exp(x)*(1 + 11*x + 14*x^2 + 3*x^3). (End)